我知道这和刚才那个人问的完全不同。解决这个问题的另一种方法是反思。我的意思是，这并没有从泛型中获益，但它让你在某种程度上模拟你想要执行的行为(让狗叫，让鸭子嘎嘎叫，等等)，而不考虑类型转换:

import java.lang.reflect.InvocationTargetException;
import java.util.HashMap;
import java.util.Map;

abstract class AnimalExample {
    private Map<String,Class<?>> friends = new HashMap<String,Class<?>>();
    private Map<String,Object> theFriends = new HashMap<String,Object>();

    public void addFriend(String name, Object friend){
        friends.put(name,friend.getClass());
        theFriends.put(name, friend);
    }

    public void makeMyFriendSpeak(String name){
        try {
            friends.get(name).getMethod("speak").invoke(theFriends.get(name));
        } catch (IllegalArgumentException e) {
            e.printStackTrace();
        } catch (SecurityException e) {
            e.printStackTrace();
        } catch (IllegalAccessException e) {
            e.printStackTrace();
        } catch (InvocationTargetException e) {
            e.printStackTrace();
        } catch (NoSuchMethodException e) {
            e.printStackTrace();
        }
    } 

    public abstract void speak ();
};

class Dog extends Animal {
    public void speak () {
        System.out.println("woof!");
    }
}

class Duck extends Animal {
    public void speak () {
        System.out.println("quack!");
    }
}

class Cat extends Animal {
    public void speak () {
        System.out.println("miauu!");
    }
}

public class AnimalExample {

    public static void main (String [] args) {

        Cat felix = new Cat ();
        felix.addFriend("Spike", new Dog());
        felix.addFriend("Donald", new Duck());
        felix.makeMyFriendSpeak("Spike");
        felix.makeMyFriendSpeak("Donald");

    }

}

由于这个问题是基于假设的数据，这里有一个很好的例子，返回一个扩展Comparable接口的泛型。

public class MaximumTest {
    // find the max value using Comparable interface
    public static <T extends Comparable<T>> T maximum(T x, T y, T z) {
        T max = x; // assume that x is initially the largest

        if (y.compareTo(max) > 0){
            max = y; // y is the large now
        }
        if (z.compareTo(max) > 0){
            max = z; // z is the large now
        }
        return max; // returns the maximum value
    }    


    //testing with an ordinary main method
    public static void main(String args[]) {
        System.out.printf("Maximum of %d, %d and %d is %d\n\n", 3, 4, 5, maximum(3, 4, 5));
        System.out.printf("Maximum of %.1f, %.1f and %.1f is %.1f\n\n", 6.6, 8.8, 7.7, maximum(6.6, 8.8, 7.7));
        System.out.printf("Maximum of %s, %s and %s is %s\n", "strawberry", "apple", "orange",
                maximum("strawberry", "apple", "orange"));
    }
}

不可能的。如果只给一个String键，Map怎么知道它将得到Animal的哪个子类呢?

唯一可能的方法是每个Animal只接受一种类型的friend(那么它可以是Animal类的一个参数)，或者callFriend()方法获得了一个类型参数。但是，看起来您确实忽略了继承的要点:在独占地使用超类方法时，只能统一地对待子类。

这个问题与Effective Java中的第29项非常相似——“考虑类型安全的异构容器”。拉兹的答案最接近布洛赫的答案。但是，为了安全起见，put和get都应该使用Class字面值。签名将变成:

public <T extends Animal> void addFriend(String name, Class<T> type, T animal);
public <T extends Animal> T callFriend(String name, Class<T> type);

在这两个方法中，您应该检查参数是否正常。有关更多信息，请参阅有效Java和类javadoc。

是什么

public class Animal {
    private Map<String,<T extends Animal>> friends = new HashMap<String,<T extends Animal>>();

    public <T extends Animal> void addFriend(String name, T animal){
        friends.put(name,animal);
    }

    public <T extends Animal> T callFriend(String name){
        return friends.get(name);
    }
}

下面是一个简单的版本:

public <T> T callFriend(String name) {
    return (T) friends.get(name); //Casting to T not needed in this case but its a good practice to do
}

完全工作的代码:

    public class Test {
        public static class Animal {
            private Map<String,Animal> friends = new HashMap<>();

            public void addFriend(String name, Animal animal){
                friends.put(name,animal);
            }

            public <T> T callFriend(String name){
                return (T) friends.get(name);
            }
        }

        public static class Dog extends Animal {

            public void bark() {
                System.out.println("i am dog");
            }
        }

        public static class Duck extends Animal {

            public void quack() {
                System.out.println("i am duck");
            }
        }

        public static void main(String [] args) {
            Animal animals = new Animal();
            animals.addFriend("dog", new Dog());
            animals.addFriend("duck", new Duck());

            Dog dog = animals.callFriend("dog");
            dog.bark();

            Duck duck = animals.callFriend("duck");
            duck.quack();

        }
    }